Đặt   \(A=\frac{1}{3}-\frac{2}{3^2}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow3A=1-\frac{2}{3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

\(4A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt    \(B=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)

\(\Rightarrow3B=3+1+...+\frac{3}{3^{98}}\)

\(2B=3-\frac{1}{3^{99}}\)

\(B=\frac{3}{2}-\frac{1}{3^{99}.2}\)

Thay B vào 4A ta có:

\(4A=\frac{3}{2}-\frac{1}{3^{99}.2}\)

\(A=\frac{3}{2.4}-\frac{1}{3^{99}.2.4}\)

\(A=\frac{3}{8}-\frac{1}{3^{99}.8}\)

Vì \(\frac{3}{8}>\frac{3}{16}\)

\(\Rightarrow\frac{3}{8}-\frac{1}{3^{99}.8}< \frac{3}{16}\)

Vậy \(A< \frac{3}{16}\)

Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)

\(3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)

\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)

\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(6A-2A=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)

\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)

\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)

\(4A=3-\frac{203}{3^{100}}< 3\)

\(A< \frac{3}{4}\left(đpcm\right)\)

CMR: \(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{100}{4^{100}}< \frac{4}{9}\)

Dạng tổng quát: CMR: \(\frac{1}{k}+\frac{2}{k^2}+\frac{3}{k^3}+\frac{4}{k^4}+...+\frac{n}{k^n}< \frac{k}{\left(k-1\right)^2}\)(k;n \(\in\) N*; k > 1)

ta có: \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{100^2}=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)

Lại có: \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};\frac{1}{4^2}>\frac{1}{4.5};...;\frac{1}{100^2}>\frac{1}{100.101}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\)

                                                                               \(=\frac{1}{2}-\frac{1}{101}\)

\(\Rightarrow1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>1-\left(\frac{1}{2}-\frac{1}{101}\right)=1-\frac{1}{2}+\frac{1}{101}\)

                                                                                                                                 \(=\frac{1}{2}+\frac{1}{101}\)

mà \(\frac{1}{2}=\frac{50}{100}>\frac{1}{100}\Rightarrow\frac{1}{2}+\frac{1}{101}>\frac{1}{100}\)

=> đ p c m

Bài 1:
Ta có:

\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)

\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Mà \(\frac{99}{100}< 1\)

\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)

Có phải ở sách NCPT ko bn

Bài 2: Đặt \(B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)

\(3B=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)

\(3B-B=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)

\(2B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(6B=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(6B-2B=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)

\(4B=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)

\(4B=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)

\(4B=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)

\(4B=3-\frac{203}{3^{100}}< 3\)

\(B< \frac{3}{4}\left(đpcm\right)\)

\(A=\frac{3}{1}+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+3+...+99+100}\)

\(=3+\frac{3}{\frac{\left(1+2\right).2}{2}}+\frac{3}{\frac{\left(1+3\right).3}{2}}+...+\frac{3}{\frac{\left(1+100\right).100}{2}}\)

\(=3+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}=3+6.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)\)

\(=3+6.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)

\(=3+6.\left(\frac{1}{2}-\frac{1}{101}\right)=3+6.\frac{99}{202}=\frac{600}{101}\)

Tốt nhất bạn nên nói mấy bài đơn giản ik dạng nâng cao ko có cho thi đâu đừng lo

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